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2 years ago

What line separates two lanes traveling in the same direction

Engineering

2 answers:

soldier1979 [14.2K]2 years ago

5 0

Answer:

White lane lines separate lanes of traffic moving in the same direction. (UK)

wolverine [178]2 years ago

3 0

In the US, we tend to have white lines separating lanes that go in the same direction, & yellow lines that separate opposing traffic.

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Need help with these 3 ez questions pls help me.

vivado [14]

Answer:

21: False 22: Unprofessional behavior 23: True Explanation:

Hope this helps!

7 0

3 years ago

1) Each of the following would be considered company-confidential except

AysviL [449]

Answer is your company’s address

8 0

3 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

Does the location of a millimeter change the voltage or current of the circuit?

Cerrena [4.2K]

Answer:

Yes, it does.

Explanation:

8 0

3 years ago

PythonA group of statisticians at a local college has asked you to create a set of functionsthat compute the median and mode of

skelet666 [1.2K]

Answer:

def median(l):
if(len(l) == 0):
return 0
else:
l.sort()
if(len(l)%2 == 0):
index = int(len(l)/2)
mid = (l[index-1] + l[index]) / 2
else:
mid = l[len(l)//2]
return mid
def mode(l):
if(len(l)==0):
return 0
mode = max(set(l), key=l.count)
return mode
def mean(l):
if(len(l)==0):
return 0
sum = 0
for x in l:
sum += x
mean = sum / len(l)
return mean
lst = [5, 7, 10, 11, 12, 12, 13, 15, 25, 30, 45, 61]
print(mean(lst))
print(median(lst))
print(mode(lst))

Explanation:

Firstly, we create a median function (Line 1). This function will check if the the length of list is zero and also if it is an even number. If the length is zero (empty list), it return zero (Line 2-3). If it is an even number, it will calculate the median by summing up two middle index values and divide them by two (Line 6-8). Or if the length is an odd, it will simply take the middle index value and return it as output (Line 9-10).

In mode function, after checking the length of list, we use the max function to estimate the maximum count of the item in list (Line 17) and use it as mode.

In mean function, after checking the length of list, we create a sum variable and then use a loop to add the item of list to sum (Line 23-25). After the loop, divide sum by the length of list to get the mean (Line 26).

In the main program, we test the three functions using a sample list and we shall get

20.5

12.5

3 0

3 years ago